It is known that the image of quadratic function y = ax05 + BX + C passes through points a (3,0), B (2, - 3), C (0. - 3) (1) The analytic formula and the axis of symmetry have been worked out as y = x & # 178; - 2x-3 and x = 1 respectively (2) Point P starts from point B and moves along the line BC to point C at the speed of 0.1 unit per second. Point Q starts from point O and moves at the same speed The velocity moves along the line OA to point A. when one of the moving points reaches the end point, the other stops moving. Let the moving time be T seconds 1. What is the value of T when the quadrilateral abpq is an isosceles trapezoid, 2. Let the intersection of PQ and the axis of symmetry be m, let the parallel line passing through M be the intersection of X axis and ab at n, and let the area of the quadrilateral anpq be s , find the analytic expression of the function of area s with respect to time t, and point out the value range of T,

It is known that the image of quadratic function y = ax05 + BX + C passes through points a (3,0), B (2, - 3), C (0. - 3) (1) The analytic formula and the axis of symmetry have been worked out as y = x & # 178; - 2x-3 and x = 1 respectively (2) Point P starts from point B and moves along the line BC to point C at the speed of 0.1 unit per second. Point Q starts from point O and moves at the same speed The velocity moves along the line OA to point A. when one of the moving points reaches the end point, the other stops moving. Let the moving time be T seconds 1. What is the value of T when the quadrilateral abpq is an isosceles trapezoid, 2. Let the intersection of PQ and the axis of symmetry be m, let the parallel line passing through M be the intersection of X axis and ab at n, and let the area of the quadrilateral anpq be s , find the analytic expression of the function of area s with respect to time t, and point out the value range of T,

1) When 0 ＜ t ＜ 10, that is, q is on the left side of the axis of symmetry, P is on the right side of the axis of symmetry, let the axis of symmetry intersect with OA and F, BC and E, then QF = 1-0,1t, EP = be-bp = 1-0.1t, so QF = EP is easy to prove △ QFM ≌ △ MEP, so FM = EM = 1 / 2Fe = 1 / 2 * 3 = 1.5, and y = - 1.5 is substituted into y = x & # 178; - 2x-3, then the sit of N can be obtained

Given the quadratic function y = f (x) = ax square + BX + C, the image passes through point a (0,2)
And f (- 1) = f (1), f (2) = 2F (1), find the quadratic function

If f (0) = C = 2 and f (- 1) = f (1) f (2) = 2F (1), then f (- 1) = A-B + C = A-B + 2F (1) = a + B + C = a + B + 2F (- 1) = f (1), A-B + 2 = a + B + 2, B = 0f (2) = 2F (1) 4A + 2B + C = 2 (a + B + C) 4A + 2 = 2A + 4, a = 1, B = 0, C = 2F (x) = x ^ 2 + 2

It is proved that the two zeros of the quadratic function FX = ax + BX + C are on both sides of the point (m, 0) if and only if AF (m)

Just say that M is between two
Let f (x) be x1, X2 and X10 P respectively
(3)af(m)

Given that the quadratic function f (x) = ax & sup2; + BX + C, and for any x ∈ R, 2aX + B = f (x + 1) + X & sup2; holds, the analytic expression of F (x) is obtained

It has been known that:
F(x+1)+x^2=a(x+1)^2+b(x+1)+c+x^2=(a+1)x^2+(2a+b)x+a+b+c (1)
It is also known that for any x ∈ R, 2aX + B = f (x + 1) + x ^ 2 is constant
a+1=0;
2a=2a+b;
b=a+b+c;
The solution is a = - 1, B = 0, C = 1
So f (x) = - x ^ 2 + 1

The math problem of senior one is about function
Given that f (x) is an even function, G (x) is an odd function, and f (x) + G (x) = 1 / X-1, the analytic expressions of F (x), G (x) are obtained
Why is - x equal to - 1 / X - 1

f(-x) =f(x)
g(-x)=-g(x)
So f (- x) + G (- x) = f (x) - G (x) = - 1 / X-1
f(x)+g(x) =1/x-1
Add two formulas
2f(x)=-2 ,f(x)=-1
So g (x) = 1 / X
And G (x) is an odd function
So g (x) is a piecewise function
g(x)=
{1/x x≠0
{0 x=0

Given that the vector M = (2Sin (Wx + π / 3), 1), the vector n = (2coswx, - √ 3), the distance between two adjacent symmetric axes of the function f (x) = m × n is π / 2 (W > 0)
When x ∈ closed interval [- 5 π / 6, π / 12], find the range of F (x)

1
f(x)=m·n=(2sin(wx+π/3),1)·(2cos(wx),-√3)
=4sin(wx+π/3)cos(wx)-√3
=4(sin(wx)/2+√3cos(wx)/2)cos(wx)-√3
=sin(2wx)+√3(1+cos(2wx))-√3
=sin(2wx)+√3cos(2wx)
=2sin(2wx+π/3)
The distance between adjacent symmetry axes is π / 2, that is, the minimum positive period: T = π
That is: 2 π / (2W) = π, that is: w = 1
That is: F (x) = 2Sin (2x + π / 3)
Increasing interval: 2x + π / 3 ∈ [2K π - π / 2,2k π + π / 2]
That is: X ∈ [K π - 5 π / 12, K π + π / 12], K ∈ Z
two
x∈[-5π/6,π/12],2x+π/3∈[-4π/3,π/2]
Sin (2x + π / 3) ∈ [- 1,1], that is: 2Sin (2x + π / 3) ∈ [- 2,2]
That is: F (x) ∈ [- 2,2]

Given the function f (x) = 2Sin (Wx - π / 6) (W > 0), the distance between two adjacent symmetry axes of an image is π / 2
(1) Finding monotone increasing interval of function;
(2) Find the range of function in the interval [0,3 π / 4]

(1) The distance between two adjacent symmetry axes of F (x) is π / 2. T / 2 = π / 2, so t = π, t = 2 π / w = π, w = 2F (x) = 2Sin (2x - π / 6) - π / 2 + 2K π ≤ 2x - π / 6 ≤ π / 2 + 2K π - π / 6 + K π ≤ x ≤ π / 3 + K π, so the monotone increasing interval is [- π / 6 + K π, π / 3 + K π], K ∈ Z (2) 0 ≤ x ≤

The function f (x) = sin (ω x + π / 6) is known, where ω > 0. If the distance between two adjacent symmetry axes of the image of function f (x) is equal to π / 2
If the equation f (x + π / 6) + mcosx + 3 = couple of X has a real number solution in X ∈ (0, π / 2), find the value range of real number M
When I use M = - 2 (cosx + 1 / cosx), the answer is m < - 4,
Correction: F (x + π / 6) + mcosx + 3 = 0,

What's your equation about x? What does it mean that f (x + π / 6) + mcosx + 3 = even has real solution in X ∈ (0, π / 2)
From your promotion, we can get ω = 2. As for the following, please explain the title and requirements in detail

Given that the function f (x) = sin (Wx + a) (W > 0, 0, less than or equal to a, less than or equal to pie) is an even number, the distance between two adjacent symmetrical axes of an image is the value of pie to find w and a
B

Even function, f (x) = f (- x)
sin(wx+a)=sin(-wx+a)
sinwxcosa+coswxsina=sin(-wx)cosa+cos(-wx)sina
sinwxcosa=-sinwxcosa
cosa=0,a=π/2
The interval between adjacent symmetrical axes is π, and the period is 2 π, so w = 1

Let f (x) = 1-2x / 1 + X, the image of y = g (x) and the image of y = f (x) be symmetric with respect to the line y = x, then G (1) =?

The image of the function y = g (x) and the image of y = f (x) are symmetric with respect to the line y = x, which shows that y = g (x) and y = f (x) are inverse functions,
Method 1: G (1) is required. Let f (x) = 1-2x / 1 + x = 0, then G (1) = 0
Method 2: find out the inverse function of function f (x) = 1-2x / 1 + X, y = 1-1x / 2 + X, that is y = g (x) = 1-1x / 2 + X
If x = 1, G (1) = 0
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